wrote in message
ups.com...



I can send you a JPEG of the circuit board if it can help, but below
is a brief description of the connections:

A picture is worth a thousand words... just remove the word "delete" from my
email address.



If so, I can't
quite understand why it is rectified prior to the transformer.


Nor can I. The transformer would only produce the other half of the wave on
the secondaries. This can't be right.


Yes, at present I have disconnected all connectors (motor, pump etc.)
from the circuit board.

Thanks for the feedback. How do I actually test the pump? Should I
fill the washing machine with water before testing the pump and see if
water exits the drain? Or is it possible to test the pump without
fluid? Anyway, at present the pump is not connected to the circuit
board.


Test the pump wiring harness. Look for a short.

Assuming the pump to operate at mains voltage, just figure out which pins on
the pump connector are for power, connect up a reasonable-guage power cord
and plug it in. We're not trying to determine if this thing pumps water...
at this point we don't really care. We just want to make sure that the
windings aren't shorted and that the pump will start and run. It won't hurt
it to run it dry.




If the triac fail shorted it could explain the excessive current
drain?

That was my original thought, but I had some reservations, hence my
questions about the existence of a relay which you have confirmed. The pump
only operates at one speed, so can be controlled using a relay as opposed to
a triac. The triac likely controls the motor so you can have slow-speed
agitation and high-speed spin with no mechanical transmission. IF the triac
controlled the pump and IF the triac were shorted, the pump would simply run
all the time. This is not the case.


The triac is connected to the circuit board via a large heatsink, so
it seems difficult to remove (hence test for shorts). Any suggestions?

Confirm that the wash/spin motor works and that it's controlled by the large
triac and move on.


If it's a relay and not a triac which controls the pump I'd take a good
look
at it... are the contacts burnt? What else is around it in the circuit?


When I trace the 3pin connector going to the pump back to the circuit
board, it seems that a relay (10A 125V AC, 12VDC) controls the pump.

Visual inspection of the circuit board indicates damage to resistor
R1. Black soot is seen in the vicinity of diodes D1, D2, D3. The
diodes does not seem damaged from visual inspection, although the
wires of the diodes are not shiny, indicating current drain.

Are diodes D1/D2/D3 and resistor R1 on the controlling 12VDC side of the
relay or the load side?

If they're on the load side, and somehow I doubt they are, it's gotta be the
pump or wiring. If they're on the control side, look for a smoked resistor
(and perhaps diode) which limits current through the relay coil. If this
failed it may take down any number of other components in the relay control
circuit.


Sorry, it is actually TOP209P. At the top is written K021 and at
bottom 1067 1B,

Well well well, that's a 3-terminal off-line PWM switch, as predicted. How
many windings/leads on the transformer. The typical circuit diagram in the
TOP209 datasheet shows it being used to derive a stable DC voltage via a
transformer with a bias winding. It's possible that this guy produces the
12VDC to operate the relay. A schematic would be really helpful here.


Thanks for all your kind help.
Best regards,
Theo

Dave

Best regards,
Theo van der Merwe

There's certainly a shorted component which burned up the resistor.
Resistors don't just burn up on their own.

Thanks for the email message and feedback. The problem occurred when
testing the pumping out of water (which wasn't working at all -
everything else seems OK). However, I have disconnected all the
external connections to the motor and pump from the circuit board. Any
suggestions on how to proceed with fault finding (assuming some
component is shorted on the circuit board)? At present I don't
understand the power supply design on the circuit board.

Best regards,
Theo van der Merwe